Increase in temperature of water = 0.53 °C
Explanation:
Change in mechanical energy = Potential energy
Potential energy = mgh
Mass, m = Mass of 1 L water = 1 kg
Acceleration due to gravity, g = 9.81 m/s²
Height, h = 225 m
Potential energy = 1 x 9.81 x 225 = 2207.25 J
Because of this 2207.25 J water gets heated.
Heat energy, E = mcΔT
Mass, m = Mass of 1 L water = 1 kg
Specific heat of water, c = 4200 J/kg/C
Energy, E = 2207.25 J
Change in temperature, ΔT = ?
Substituting
2207.25 = 1 x 4200 x ΔT
ΔT = 0.53 °C
Increase in temperature of water = 0.53 °C
Answer:
force
Explanation:
If you need explanation lmk and if that's not the answer lmk as well so i can think of other one's hope it helps
Using safe research practices would be a very good common-sense practice in a lab environment.
<span>d. people marketing
</span>Many firms develop a formal _____ to answer the question: who is your target market and how do you plan to reach them? people marketing
NOT:
a. marketing mix
b. market vision statement
<span>c. marketing priority list</span>
Answer:
θ = 6.3 *10³ revolutions
Explanation:
Angular acceleration of the drill
We apply the equations of circular motion uniformly accelerated
ωf= ω₀ + α*t Formula (1)
Where:
α : Angular acceleration (rad/s²)
ω₀ : Initial angular speed ( rad/s)
ωf : Final angular speed ( rad
t : time interval (s)
Data
ω₀ = 0
ωf = 350000 rpm = 350000 rev/min
1 rev = 2π rad
1 min= 60 s
ωf = 350000 rev/min =350000*(2π rad/60 s)
ωf = 36651.9 rad/s
t = 2.2 s
We replace data in the formula (2) :
ωf= ω₀ + α*t
36651.9 = 0 + α* (2.2)
α = 36651.9 / (2.2)
α = 17000 rad/s²
Revolutions made by the drill
We apply the equations of circular motion uniformly accelerated
ωf²= ω₀ ²+ 2α*θ Formula (2)
Where:
θ : Angle that the body has rotated in a given time interval (rad)
We replace data in the formula (2):
(ωf)²= ω₀²+ 2α*θ
(36651.9)²= (0)²+ 2( 17000 )*θ
θ = (36651.9)²/ (34000 )
θ = 39510.64 rad = 39510.64 rad* (1 rev/2πrad)
θ = 6288.31 revolutions
θ = 6.3 *10³ revolutions
